题目描述:
Given a binary search tree, write a function kthSmallest
to find the kth smallest element in it.
Note:
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.
Example 1:
Input: root = [3,1,4,null,2], k = 1
3
/ \
1 4
\
2
Output: 1
Example 2:
Input: root = [5,3,6,2,4,null,null,1], k = 3
5
/ \
3 6
/ \
2 4
/
1
Output: 3
Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?
求BST的第k小的值,先计算左子树的节点数,如果数量大于等于k说明目标在左子树,如果等于k-1,说明根节点即为所求,否则说明目标值在右子树。
class Solution {
public:
int kthSmallest(TreeNode* root, int k) {
int left_num=countNode(root->left);
if(left_num==k-1) return root->val;
else if(left_num>=k) return kthSmallest(root->left,k);
else if(left_num<k-1) return kthSmallest(root->right,k-left_num-1);
}
int countNode(TreeNode* root)
{
if(root==NULL) return 0;
return 1+countNode(root->left)+countNode(root->right);
}
};